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Explanation:

If we split the alphabets in a group of 3 we get following arrangement:

D__F | __DE | E__D | __EF | __DE | __F

Now, if we put option 3, we get:

DEF | FDE | EFD | DEF | FDE | EF

If we see the pattern it is consistent in the sense that the third alphabet in each subset is the first alphabet in the next subset and the order is cyclic. (D, E, F or F, D, E or E, F, D).

Hence, option (c).

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